# Edexcel M4 Exam Discussion Thread [June 5th] Watch

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#121

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The one where you are required to find the speed of B.

**simonli2575**)The one where you are required to find the speed of B.

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If it is then I got that question completely wrong. Think I dropped all the marks but 1. Still ended up with an A* on the paper though because the boundary was so low

**Gome44**)If it is then I got that question completely wrong. Think I dropped all the marks but 1. Still ended up with an A* on the paper though because the boundary was so low

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#123

(Original post by

Ah no, that one was the easy one. I was talking about the one in which boat B is moving due east and some time later, B is 2sqrt(3) km due south of A.

**simonli2575**)Ah no, that one was the easy one. I was talking about the one in which boat B is moving due east and some time later, B is 2sqrt(3) km due south of A.

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#124

(Original post by

Me too(M3 90,Fp2 100). How did you do in FP2, M3? M4 is not the best unit to rely on (you can aim for A* but not be sure)

If you have M2 and/or FP3 it is better to think of getting a good mark in these units,in my opinion( they are safer)

**Vesniep**)Me too(M3 90,Fp2 100). How did you do in FP2, M3? M4 is not the best unit to rely on (you can aim for A* but not be sure)

If you have M2 and/or FP3 it is better to think of getting a good mark in these units,in my opinion( they are safer)

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#125

Hoping for a paper like last years (mostly easy but some tricky parts so the grade boundaries fall)

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#126

What is the reason for why the component of the velocity that is perpendicular to the common remains unchanged? Also, why does the fact that all collisions in M4 are smooth mean that the mutual reaction acts along the common normal at the point of impact?

If we were solving a problem involving collisions of rough elastic bodies, what would change?

If we were solving a problem involving collisions of rough elastic bodies, what would change?

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#127

(Original post by

What is the reason for why the component of the velocity that is perpendicular to the common remains unchanged? Also, why does the fact that all collisions in M4 are smooth mean that the mutual reaction acts along the common normal at the point of impact?

If we were solving a problem involving collisions of rough elastic bodies, what would change?

**ThatPerson**)What is the reason for why the component of the velocity that is perpendicular to the common remains unchanged? Also, why does the fact that all collisions in M4 are smooth mean that the mutual reaction acts along the common normal at the point of impact?

If we were solving a problem involving collisions of rough elastic bodies, what would change?

This means an impulse is only felt parallel to the line of centres and therefore the velocities perpendicular to this line are unchanged (impulse is 0).

If the objects are completely smooth then there is no force perpendicular to the normal. If they were rough there would be a component frictional force perpendicular to the normal and so both components of velocity would change rather than just one.

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#128

(Original post by

When two objects collide, they exert a force upon each other perpendicular to their plane of contact (along the normal of the plane).

This means an impulse is only felt parallel to the line of centres and therefore the velocities perpendicular to this line are unchanged (impulse is 0).

If the objects are completely smooth then there is no force perpendicular to the normal. If they were rough there would be a component frictional force perpendicular to the normal and so both components of velocity would change rather than just one.

**Gawain**)When two objects collide, they exert a force upon each other perpendicular to their plane of contact (along the normal of the plane).

This means an impulse is only felt parallel to the line of centres and therefore the velocities perpendicular to this line are unchanged (impulse is 0).

If the objects are completely smooth then there is no force perpendicular to the normal. If they were rough there would be a component frictional force perpendicular to the normal and so both components of velocity would change rather than just one.

"A small smooth spherical ball of mass m falls vertically at strikes a fixed smooth inclined plane with speed u.

(a) Explain why the component of the velocity of the ball parallel to the plane is not affected by the impact"

The answer would be something like "When the collision occurs the ball and plane exert a force on each other that is perpendicular to the plane, hence there is no impulse parallel to the plane and so the parallel velocity is unchanged"?

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#129

**quick question on the relative motion (M4 edexcel)**

only a few minutes old ..

I will post the solution real soon

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(Original post by

only a few minutes old ..

I will post the solution real soon

**TeeEm**)**quick question on the relative motion (M4 edexcel)**only a few minutes old ..

I will post the solution real soon

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#131

(Original post by

How about a question using relative velocities to find the minimum separation given A is trying go get away from B?

**simonli2575**)How about a question using relative velocities to find the minimum separation given A is trying go get away from B?

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#132

**TeeEm**)

**quick question on the relative motion (M4 edexcel)**

only a few minutes old ..

I will post the solution real soon

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#133

Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??

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#134

Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5

Posted from TSR Mobile

Posted from TSR Mobile

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#135

(Original post by

Thanks for the explanation. So if I have the following question:

"A small smooth spherical ball of mass m falls vertically at strikes a fixed smooth inclined plane with speed u.

(a) Explain why the component of the velocity of the ball parallel to the plane is not affected by the impact"

The answer would be something like "When the collision occurs the ball and plane exert a force on each other that is perpendicular to the plane, hence there is no impulse parallel to the plane and so the parallel velocity is unchanged"?

**ThatPerson**)Thanks for the explanation. So if I have the following question:

"A small smooth spherical ball of mass m falls vertically at strikes a fixed smooth inclined plane with speed u.

(a) Explain why the component of the velocity of the ball parallel to the plane is not affected by the impact"

The answer would be something like "When the collision occurs the ball and plane exert a force on each other that is perpendicular to the plane, hence there is no impulse parallel to the plane and so the parallel velocity is unchanged"?

(Original post by

Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5

Posted from TSR Mobile

**CowsWhoStare**)Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5

Posted from TSR Mobile

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).

(Original post by

Question with solution from post 129

**TeeEm**)Question with solution from post 129

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#136

(Original post by

Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??

**jjpneed1**)Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??

If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).

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#137

(Original post by

Yeah that looks good.

To find the minimum velocity for interception, the relative velocity (which is in the direction of the initial distance) is perpendicular to the 'interceptor's' velocity.

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).

This is just a stylistic question but do you think it is better to separate your diagrams or combine then like this:

**Gawain**)Yeah that looks good.

To find the minimum velocity for interception, the relative velocity (which is in the direction of the initial distance) is perpendicular to the 'interceptor's' velocity.

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).

This is just a stylistic question but do you think it is better to separate your diagrams or combine then like this:

Apologies to any student which finds my style difficult to follow, however feel free to attach an alternative solution if it is going to benefit some students

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#138

(Original post by

everybody has their style and this particular type of question has way too many styles ...

Apologies to any student which finds my style difficult to follow, however feel free to attach an alternative solution if it is going to benefit some students

**TeeEm**)everybody has their style and this particular type of question has way too many styles ...

Apologies to any student which finds my style difficult to follow, however feel free to attach an alternative solution if it is going to benefit some students

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#139

(Original post by

I don't find your style confusing, to the contrary I think it's clearer. It's just mark schemes combine there diagrams so I thought that's what is expected.

**Gawain**)I don't find your style confusing, to the contrary I think it's clearer. It's just mark schemes combine there diagrams so I thought that's what is expected.

The way I present it is the way I understood the topic as a learner a very long time ago. (Fashions do change)

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#140

(Original post by

No that's right,

If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).

**Gawain**)No that's right,

If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).

Also another problem with Q5 on the same paper

In the conservation of momentum calculation I used x and y going in the opposite direction to that shown in the diagram in the mark scheme, so

-u = x + 3y and x - y = u/3

but this gives x=0, y=-u/3 which is wrong... why would it matter which why I label the velocities...? Somehow I went through the other mechanics modules without encountering this problem

EDIT: nvm sign error using restitution I believe

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